From Linear Dynamics to Convolutions
A linear stochastic differential equation with additive noise has a useful structure: its drift can amplify or damp earlier inputs, while its diffusion adds random fluctuations over time. The effect at time \(t\) of an input applied at an earlier time \(s\) is weighted by a deterministic factor. Integrating these weighted inputs produces a deterministic convolution for the forcing and a stochastic convolution for the noise.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Fix a finite horizon \(T\), a deterministic initial value \(x_0\), and deterministic measurable coefficient functions \(a_t,f_t,g_t\) on \([0,T]\). Assume \(a\) and \(f\) are Lebesgue integrable and \(g\) is square-integrable on \([0,T]\). These conditions ensure the ordinary integrals below exist and, on a finite horizon, the deterministic integrands in the stochastic integrals are square-integrable.
All stochastic integrals in this tutorial are Itô integrals. For clarity, an integral \(\int_0^t H_s\,dB_s\) is taken with an adapted, progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); this class is denoted \(\mathcal{L}^2\). The deterministic integrands used below are progressively measurable and belong to \(\mathcal{L}^2\). We use the Itô multiplication rules \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\), not the ordinary chain rule. No Stratonovich integral is being used.
The Convolution Formula
Proof. Since \(a\) is integrable, \(A\) is absolutely continuous and \(e^{-A(t)}\) has finite variation. Apply Itô’s formula to \(F(t,x)=e^{-A(t)}x\). Its derivatives are \(F_t(t,x)=-a_t e^{-A(t)}x\), \(F_x(t,x)=e^{-A(t)}\), and \(F_{xx}(t,x)=0\), so the Itô correction \(\tfrac12F_{xx}(t,X_t)g_t^2\,dt\) is zero. Thus, for any solution \(X\), $$ d\big(e^{-A(t)}X_t\big)=e^{-A(t)}f_t\,dt+e^{-A(t)}g_t\,dB_t. $$ Integrating from \(0\) to \(t\), then multiplying by \(e^{A(t)}\), gives the displayed formula because \(e^{A(t)-A(s)}=K(s,t)\). Conversely, the right-hand side is adapted and continuous: its ordinary integral is continuous, and its stochastic integral has a continuous version since its integrand is in \(\mathcal{L}^2\). The integrating-factor calculation therefore verifies that it satisfies the SDE. For uniqueness, let \(X\) and \(\widetilde X\) be two solutions with the same initial value and coefficients. Their difference \(D_t=X_t-\widetilde X_t\) satisfies \(dD_t=a_tD_t\,dt\) and \(D_0=0\). Applying the same integrating-factor calculation gives \(e^{-A(t)}D_t=D_0=0\), hence \(D_t=0\) for all \(t\) almost surely. \(\square\)
The stochastic convolution deserves a qualification. The process \(M_t=\int_0^t e^{-A(s)}g_s\,dB_s\) is a square-integrable martingale: its integrand is in \(\mathcal{L}^2\), and for \(u\ge t\), the Itô integral property gives \(\mathbb{E}[M_u\mid\mathcal{F}_t]=M_t\). The stochastic convolution in the solution is \(Z_t=e^{A(t)}M_t\). Since the factor \(e^{A(t)}\) changes with \(t\), \(Z_t\) is generally not a martingale.
Means and Variances
Proof. The deterministic terms in the solution are non-random. The stochastic integral has expectation zero, so taking expectations gives the stated mean. To compute the variance, subtract that mean from the solution. The difference is the Itô integral \(\int_0^t K(s,t)g_s\,dB_s\). Its integrand is in \(\mathcal{L}^2\), and the Itô isometry gives its second moment as \(\mathbb{E}\int_0^t K(s,t)^2g_s^2\,ds\). The integrand is deterministic, so this equals the displayed variance. \(\square\)
These statements concern a fixed terminal time \(t\); the deterministic kernel \(K(s,t)\) depends on that terminal time. In particular, the Itô isometry computes the variance of the convolution at \(t\), but does not say that the process of convolutions, as \(t\) varies, is a martingale.
Worked Example: Linear Growth with Constant Additive Noise
Consider \(dX_t=X_t\,dt+t\,dt+2\,dB_t\), \(X_0=1\), and find the mean and variance at \(T=\ln 2\).
Here \(a_t=1\), \(f_t=t\), and \(g_t=2\). Therefore \(A(t)=t\) and \(K(s,t)=e^{t-s}\). The mean is $$ \mathbb{E}[X_t]=e^t+e^t\int_0^t s e^{-s}\,ds =e^t+e^t\big[1-(t+1)e^{-t}\big] =2e^t-t-1. $$ At \(t=\ln 2\), \(e^t=2\), so the mean is \(4-\ln 2-1=3-\ln 2\). The variance is $$ \operatorname{Var}(X_t)=4e^{2t}\int_0^t e^{-2s}\,ds =2(e^{2t}-1). $$ At \(t=\ln 2\), this is \(2(4-1)=6\).
Check by substitution: The formula gives \(X_t=e^t\big(1+\int_0^t s e^{-s}\,ds+\int_0^t 2e^{-s}\,dB_s\big)\). Write the expression in parentheses as \(Y_t\). Then \(dY_t=t e^{-t}\,dt+2e^{-t}\,dB_t\). Apply Itô’s formula to \(F(t,y)=e^t y\): \(F_t=e^t y\), \(F_y=e^t\), \(F_{yy}=0\), so the correction \(\tfrac12F_{yy}(2e^{-t})^2dt\) is zero. It follows that \(dX_t=X_t\,dt+t\,dt+2\,dB_t\), as required. Also \(X_0=1\), confirming the initial condition.
Worked Example: Time-Varying Drift and Forcing
Let \(dX_t=2tX_t\,dt+e^{t^2}\,dt+e^{t^2}\,dB_t\), \(X_0=0\). Compute its mean and variance at \(t=1\).
Now \(A(t)=\int_0^t2r\,dr=t^2\), so \(K(s,t)=e^{t^2-s^2}\). The deterministic convolution has integrand \(K(s,t)f_s=e^{t^2-s^2}e^{s^2}=e^{t^2}\), hence $$ \mathbb{E}[X_t]=\int_0^t e^{t^2}\,ds=t e^{t^2}. $$ The stochastic convolution has integrand \(K(s,t)g_s=e^{t^2}\), and therefore $$ \operatorname{Var}(X_t)=\int_0^t e^{2t^2}\,ds=t e^{2t^2}. $$ At \(t=1\), the mean is \(e\) and the variance is \(e^2\).
Check by substitution: The formula reduces to \(X_t=e^{t^2}(t+B_t)\). For \(F(t,b)=e^{t^2}(t+b)\), the derivatives are \(F_t=e^{t^2}[1+2t(t+b)]\), \(F_b=e^{t^2}\), and \(F_{bb}=0\). Thus the Itô correction \(\tfrac12F_{bb}\,dt\) is zero, and \(dX_t=[e^{t^2}+2tX_t]\,dt+e^{t^2}\,dB_t\). This is exactly the given equation, and \(X_0=0\).
Worked Example: A Time-Dependent Damping Coefficient
Consider \(dX_t=-2tX_t\,dt+e^{-t^2}\,dB_t\) with \(X_0=2\). Find the mean and variance at \(t=1\).
Here \(A(t)=-t^2\), so \(K(s,t)=e^{-t^2+s^2}\). Since \(f_s=0\), the mean is \(2e^{-t^2}\). For the variance, \(K(s,t)g_s=e^{-t^2+s^2}e^{-s^2}=e^{-t^2}\), and thus $$ \operatorname{Var}(X_t)=\int_0^t e^{-2t^2}\,ds=t e^{-2t^2}. $$ At \(t=1\), the mean is \(2e^{-1}\) and the variance is \(e^{-2}\).
Check by substitution: The formula gives \(X_t=e^{-t^2}(2+B_t)\). For \(F(t,b)=e^{-t^2}(2+b)\), \(F_t=-2t e^{-t^2}(2+b)\), \(F_b=e^{-t^2}\), and \(F_{bb}=0\). The Itô correction is \(\tfrac12F_{bb}\,dt=0\), so \(dX_t=-2tX_t\,dt+e^{-t^2}\,dB_t\). At \(t=0\), this expression equals \(2\), as specified.
Why the Martingale Distinction Matters
A common error is to treat the stochastic convolution as if its changing terminal-time kernel were fixed. For example, take \(a_t=1\), \(g_t=1\), and no forcing. The stochastic convolution is \(Z_t=e^t\int_0^t e^{-s}\,dB_s\). For \(u>t\), the integral from \(t\) to \(u\) has conditional expectation zero given \(\mathcal{F}_t\), so $$ \mathbb{E}[Z_u\mid\mathcal{F}_t] =e^u\int_0^t e^{-s}\,dB_s =e^{u-t}Z_t. $$ In general this is not \(Z_t\). The incorrect calculation would call \(Z\) a martingale merely because it contains an Itô integral. The correction is to identify the actual martingale \(M_t=\int_0^t e^{-s}\,dB_s\); the convolution is \(Z_t=e^tM_t\), a time-varying multiple of that martingale.
The convolution representation is useful because it separates deterministic response from random fluctuation. It also identifies directly how earlier noise is weighted: \(K(s,t)\) describes the effect at time \(t\) of a noise input at time \(s\). The drift coefficient controls this propagation, while the additive diffusion coefficient determines the variance through its squared, propagated value. Confusing drift and diffusion, or omitting the square on \(K\) in the variance integral, changes the result.
| Quantity | Expression at time \(t\) | Role |
|---|---|---|
| Initial value contribution | \(K(0,t)x_0\) | Propagated starting state |
| Deterministic forcing | \(\int_0^t K(s,t)f_s\,ds\) | Contribution to the mean |
| Stochastic forcing | \(\int_0^t K(s,t)g_s\,dB_s\) | Zero-mean fluctuation |
| Variance | \(\int_0^t K(s,t)^2g_s^2\,ds\) | Accumulated squared noise response |
Check Your Understanding
Use the convolution formula and its stated hypotheses to answer the following.
- For \(a_t=0\), \(f_t=1\), \(g_t=3\), and \(X_0=2\), find the mean and variance at a general time \(t\ge0\).
- For deterministic coefficients, why does the stochastic convolution have mean zero even though its integrand depends on the terminal time?
- In the second worked example, recompute \(\operatorname{Var}(X_1)\) directly from the Itô isometry.
- For \(a_t=1\), \(g_t=1\), explain why \(M_t=\int_0^t e^{-s}\,dB_s\) is a martingale but \(e^tM_t\) generally is not.
- State which integrability condition on \(g\) ensures that the stochastic integrals in the solution are well-defined on a finite horizon.